Problem: There exists a positive real number $x$ such that $ \cos (\arctan (x)) = x $. Find the value of $x^2$.
Solution: Construct a right triangle with legs 1 and $x.$  Let the angle opposite the side length $x$ be $\theta.$

[asy]
unitsize(1 cm);

pair A, B, C;

A = (2,1.8);
B = (0,0);
C = (2,0);

draw(A--B--C--cycle);
draw(rightanglemark(A,C,B,8));

label("$\theta$", B + (0.7,0.3));
label("$1$", (B + C)/2, S);
label("$x$", (A + C)/2, E);
label("$\sqrt{x^2 + 1}$", (A + B)/2, NW);
[/asy]

Then $\tan \theta = x,$ so $\theta = \arctan x.$  Then
\[\cos (\arctan x) = \frac{1}{\sqrt{x^2 + 1}},\]so
\[\frac{1}{\sqrt{x^2 + 1}} = x.\]Squaring both sides, we get
\[\frac{1}{x^2 + 1} = x^2,\]so $x^4 + x^2 - 1 = 0.$  By the quadratic formula,
\[x^2 = \frac{-1 \pm \sqrt{5}}{2}.\]Since $x^2$ is positive,
\[x^2 = \boxed{\frac{-1 + \sqrt{5}}{2}}.\]